---
title: "Full-bridge inverter RLC load, THD"
subject: "Power Electronics"
chapter: 5
key: "semester_7/power_electronics:full-bridge-inverter-rlc-112uf"
source: https://www.bctengineer.com/q/semester_7/power_electronics/full-bridge-inverter-rlc-112uf
site: BCT Engineering
---

# Full-bridge inverter RLC load, THD

Power Electronics · Chapter 5: Inverter

- Asked in: 2078 Bhadra, 8 marks
- Video: [Lec 38(e) Full Bridge Inverter | RLC load | Power Electronics](https://www.youtube.com/watch?v=cc2p3yFM7Vc) — Gatematic by Sohail Sir
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/full-bridge-inverter-rlc-112uf
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5

## Question

Single phase full bridge inverter has an RLC load with R = 10Ω, L = 31.5mH and C = 112µF. The inverter frequency is 50Hz and dc input voltage is 220V. (i) express the instantaneous load current in fourier series (ii) RMS value of fundamental component of load current (iii) THD of load current

## Short answer

Use v0 as the sum of odd harmonics 4Vs over n pi sin n omega t, and divide each harmonic voltage by the series RLC impedance. Fundamental peak current is 13.31 A, so I1 rms is 9.41 A. Including harmonics up to 9th, the THD of load current is 32.9 percent.

## Answer

The square-wave output of a full-bridge inverter is $v_0 = \sum_{n\ \text{odd}}\frac{4V_s}{n\pi}\sin n\omega t$. Each harmonic current is $V_{nm}/Z_n$, with the series RLC impedance evaluated at $n\omega$.

**Given:** $R = 10\ \Omega$, $L = 31.5$ mH, $C = 112\ \mu$F, $f = 50$ Hz ($\omega = 314.16$ rad/s), $V_s = 220$ V.

$$
\begin{aligned}
V_{nm} &= \frac{4\times220}{n\pi} = \frac{280.11}{n}\ \text{V} \\
Z_n &= \sqrt{R^2 + \Big(n\omega L - \frac{1}{n\omega C}\Big)^2}, \qquad \theta_n = \tan^{-1}\frac{n\omega L - 1/(n\omega C)}{R}
\end{aligned}
$$

| $n$ | $V_{nm}$ (V) | $X_L$ (Ω) | $X_C$ (Ω) | $Z_n$ (Ω) | $\theta_n$ | $I_{nm}$ (A) |
|---|---|---|---|---|---|---|
| 1 | 280.11 | 9.90 | 28.42 | 21.05 | −61.64° | 13.31 |
| 3 | 93.37 | 29.69 | 9.47 | 22.55 | 63.68° | 4.14 |
| 5 | 56.02 | 49.48 | 5.68 | 44.92 | 77.14° | 1.25 |
| 7 | 40.02 | 69.27 | 4.06 | 65.97 | 81.28° | 0.61 |
| 9 | 31.12 | 89.06 | 3.16 | 86.49 | 83.36° | 0.36 |

#### (i) Instantaneous load current (Fourier series)
$$
\begin{aligned}
i_0(t) ={}& 13.31\sin(314.16t + 61.64^\circ) + 4.14\sin(3\omega t - 63.68^\circ) \\
&+ 1.25\sin(5\omega t - 77.14^\circ) + 0.61\sin(7\omega t - 81.28^\circ) \\
&+ 0.36\sin(9\omega t - 83.36^\circ) + \dots\ \text{A}
\end{aligned}
$$

At the fundamental the load is capacitive ($X_C > X_L$), so $i_1$ leads; at higher harmonics it is inductive, so they lag.

#### (ii) RMS of fundamental load current
$$
I_1 = \frac{13.306}{\sqrt2} = 9.41\ \text{A}
$$

#### (iii) THD of load current
RMS of each harmonic: $I_3 = 2.93$, $I_5 = 0.88$, $I_7 = 0.43$, $I_9 = 0.25$ A.

$$
\begin{aligned}
I_0 &= \sqrt{9.41^2 + 2.93^2 + 0.88^2 + 0.43^2 + 0.25^2} = 9.906\ \text{A} \\
THD &= \frac{\sqrt{I_0^2 - I_1^2}}{I_1} = \frac{\sqrt{2.93^2 + 0.88^2 + 0.43^2 + 0.25^2}}{9.41} \\
&= \frac{3.098}{9.41} = 0.329 = 32.9\%
\end{aligned}
$$

Including all higher harmonics raises it only slightly (about 33.0%).

**Answer:** $I_{1,rms} = 9.41$ A; $THD_i \approx 32.9\%$ (harmonics up to 9th).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For Vs = 220 V, what is V_nm?
   A. 220/n V
   B. (2 x 220)/(n pi) V
   C. 280.11/n V
   D. 440/n V
   Answer: C. V_nm = 4Vs/(n pi) = 280.11/n V. (Key point: Harmonic voltage amplitude)

2. Why does the fundamental current lead while higher harmonic currents lag?
   A. X_C > X_L at the fundamental, but X_L grows and X_C falls with n
   B. X_L is zero at the fundamental
   C. R changes with n
   D. The source reverses
   Answer: A. X_L = n w L rises and X_C = 1/(n w C) falls, so the load changes from capacitive to inductive. (Key point: Load nature by harmonic)

3. What is the rms fundamental current?
   A. 13.31 A
   B. 18.8 A
   C. 6.7 A
   D. 9.41 A
   Answer: D. I_1m = 280.11/21.05 = 13.31 A; rms = 13.31/sqrt2 = 9.41 A. (Key point: Fundamental rms)

4. How is current THD computed?
   A. Harmonic rms (sqrt of sum of squares of I3, I5, ...) divided by I1
   B. Sum of harmonic rms divided by I_0
   C. I1 divided by I_0
   D. Peak harmonic divided by peak I1
   Answer: A. THD = sqrt(I_0^2 - I_1^2)/I_1 = 3.098/9.41 = 32.9%. (Key point: THD)

## Related in Power Electronics

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- [Full-bridge inverter RLC load to fifth harmonic](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-bridge-inverter-rlc-155uf) — Power Electronics, chapter 5
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